Question 2 of 8: Stepped Shaft — Torsion

Question 2: Stepped Shaft — Torsion (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) the maximum shear stress in the shaft and its variation across the radius there; (b) the angle of twist at A relative to the fixed end D.

Approach. Cut each segment and sum the applied torques between the free end A and the cut to get the internal torque; compute τ=Tr/J in each segment to find the maximum, then sum φ=TL/(GJ) segment-by-segment (fixed-end convention) for the total twist at A.

1. Internal torques (CW positive, summing from the free end A inward): $$T_{AB}=1000\text{ N}\cdot\text{m},\quad T_{BC}=1000+3500=4500\text{ N}\cdot\text{m},\quad T_{CD}=4500-7000=-2500\text{ N}\cdot\text{m}$$ (CD carries 2500 N·m the opposite sense, since T_C is CCW and outweighs A+B).
2. Polar moments $J=\pi d^4/32$: $$J_{AB}=79522\text{ mm}^4,\quad J_{BC}=1272345\text{ mm}^4,\quad J_{CD}=251327\text{ mm}^4.$$
3. Maximum shear stress in each segment, $\tau=T(d/2)/J$: $$\tau_{AB}=188.6\text{ MPa},\quad \tau_{BC}=106.1\text{ MPa},\quad \tau_{CD}=198.9\text{ MPa}.$$ The largest occurs in the narrower CD segment despite carrying the smallest internal torque, because its diameter is also small: $$\boxed{\tau_{max}=198.9\text{ MPa (in CD)}}$$ — below the 220 MPa yield, so the shaft does not yield. Across the radius, τ varies linearly from 0 at the centre to this value at the outer fibre, $\tau(\rho)=\tau_{max}\rho/(d/2)$.
4. Angle of twist at A (sum of the three segment twists, each $\phi=TL/(GJ)$, algebraic sign preserved): $$\phi_A=\phi_{AB}+\phi_{BC}+\phi_{CD}=0.2841\text{ rad}$$ $$\boxed{\phi_A = 16.279^{\circ}\ \text{(same sense as the net CW torques)}}$$